More answers

next recalling that the rule for rotating 180 degrees is \((x,y)\longrightarrow (-x,-y)\) ends the proof

anonymous

2 years ago

Also consider the fact that the angle between the lines of your two reflections determines the angle of rotation (if you wanted to go that route)

anonymous

2 years ago

i.e. Two reflections are equivalent to a specific angle rotation. And the angle between the lines determines the rotation about their intersection point.

anonymous

2 years ago

Oh, and I forgot to mention that it's twice the angle between them. This means that since there is a 90 degree angle between the x- and y-axis, there will be a 180 degree rotation.

ikram002p

2 years ago

what im trying to understand is why u have chosen a regular \(2n-gon\) xD i believe it applies on each n-gon, this is confusing me since im trying to know what is special about it but in general all rotation and fillips are alike for Dihedral groups xD
or maybe i'm missing something out.
|dw:1440934919407:dw|

ganeshie8

2 years ago

@jabberwock thanks! that works perfectly!

ganeshie8

2 years ago

@ikram002p
when the number of sides is not even, we will not be haveing the element \(R_{180}\) in the dihedral group right ?

ikram002p

2 years ago

well that part is confusing me, since basically rotation in abstract algebra (R_something) mean identity and u need it to take origin form, how ever in geometry u can rotate anything u want as much as u want without such restrict xD if you know what i mean.

ikram002p

2 years ago

|dw:1440936280877:dw|

ganeshie8

2 years ago

Yes, i get what you mean.
In dihedral groups, the rotations are restricted to a finite number :
integer multiples of \(\large \dfrac{360}{n}\)

ganeshie8

2 years ago

where \(n\) = number of sides

ikram002p

2 years ago

exactly, but now i got what ur asking for xD

ganeshie8

2 years ago

after rotating, the polygon must look exactly same as before
(overlap exactly)

ganeshie8

2 years ago

that applies to reflections also
we only consider the reflections that overlap exactly with the initial polygon

ganeshie8

2 years ago

they are precisely the reflections over "symmetry lines" of the polygon

ganeshie8

2 years ago

and as you know every \(n-gon\) has \(n\) symmetry lines

ikram002p

2 years ago

it make sense to me now, been long time these stuff need to be reviewed.

anonymous

2 years ago

You can also prove this from the perspective of complex analysis. Let \(z\) and \(w\) be complex numbers such that \(z^{2n}=w\), and let \(z=r\exp\left(i\theta_0\right)\) so that \(|z|=r\) and \(\arg z=\theta_0\).
The \(2n\)th roots of \(w\) are then
\[\large w^{\frac{1}{2n}}=z=r^{\frac{1}{2n}}\exp\left(\frac{\theta_0+2\pi k}{2n}i\right),\quad k=0,1,\ldots,2n-1\]
As you probably know, the \(n\)th roots of any complex number form a regular \(n\)-gon around the origin.
A reflection across the real axis (a "horizontal flip", as you put it) is equivalent to the mapping \(\xi\mapsto\bar{\xi}\) for any \(z\) in the plane, where \(\bar{\xi}\) is the complex conjugate of \(\xi\). A reflection across the imaginary axis ("vertical flip") is the same as \(\xi\mapsto-\bar{\xi}\). Putting these transformations together yields \(\xi\mapsto-\xi\). So, any of the \(2n\)th roots \(z\) simply become \(-z\), which can be written as
\[\large -z=-r^{\frac{1}{2n}}\exp\left(\frac{\theta_0+2\pi k}{2n}i\right),\quad k=0,1,\ldots,2n-1\]
On the other hand, a rotation of \(180^\circ\), or \(\pi\text{ rad}\), can be described by writing \(\arg z\) as \(\arg z^*:=\arg z+i\pi\), so we have
\[\large z^*=r^{\frac{1}{2n}}\exp\left(\frac{\theta_0+2n\pi+2\pi k}{2n}i\right),\quad k=0,1,\ldots,2n-1\]
Now we just have to show that \(-z=z^*\):
\[\large\begin{align*}
-r^{\frac{1}{2n}}\exp\left(\frac{\theta_0+2\pi k}{2n}i\right)&=r^{\frac{1}{2n}}\exp\left(\frac{\theta_0+2n\pi+2\pi k}{2n}i\right)\\[2ex]
-\exp\left(\frac{\theta_0+2\pi k}{2n}i\right)&=\exp\left(\frac{\theta_0+(2n+2k)\pi}{2n}i\right)\end{align*}\]
Take the logarithm of both sides, using the principal branch, i.e. \(\ln \xi=\ln(se^{it})=\ln s+it\).
\[\large\begin{align*}
\ln(-1)+\frac{\theta_0+2\pi k}{2n}i&=\ln1+\frac{\theta_0+(2n+2k)\pi}{2n}i\\[2ex]
i\pi+\frac{\theta_0+2\pi k}{2n}i&=\frac{\theta_0+(2n+2k)\pi}{2n}i\end{align*}\]
and you can easily work out that the LHS=RHS.